incenter I of a triangle

The Problem:
	Recall that the incenter I of a triangle is the point where the three internal angle bisectors meet. Prove that any line through I that divides the area of the triangle in half also divides its perimeter in half; conversely, any line through I that divides the perimeter of the triangle in half also divides its area in half.

Triangle geometry employs a notation that has become standard: Our triangle ABC has sides a=BC,b=CA , and c=AB ; its incenter is I , inradius r . We use square brackets to denote area. Let ℓ be an arbitrary line through I ; because such a line must intersect two sides of ΔABC , we label the figure so that ℓ meets AC at D and AB at E . Finally, let b′=DA and c′=AE .

Our proof consists of eight equivalent statements.

ℓ divides the area ofΔ⇔⇔⇔⇔⇔⇔⇔ABC in half [ABC]=2[AED][ABI]+[BCI]+[CAI]=2([AEI]+[AID])c⋅r2+a⋅r2+b⋅r2=2(c′⋅r2+b′⋅r2)a+b+c=2(c′+b′)a+(b−b′)+(c−c′)=b′+c′BC+CD+EB=DA+AEℓ divides the perimeter of ΔABC in half.

Comments. Note that our argument remains valid should the point D coincide with the vertex C , in which case DI=CI bisects ∠ACB . As several correspondents observed, for the angle bisector CE to divide either the area or perimeter of ΔABC in half, E would necessarily be the midpoint of AB , and the triangle would be isosceles with CA=CB .

Stadler kindly sent us a reference to a "Proof Without Words" [3] which proved pictorially that a line passing through the incenter of a triangle bisects the perimeter if and only if it bisects the area. Unfortunately, that proof badly needs quite a few words to become comprehensible; even then the argument there is not as simple as what we have featured above, which is a distillation of the lovely solutions sent in by our readers.

Stronger theorems. By arguing more carefully one can prove a result appearing in [2], namely

If any two of the following statements hold about a line ℓ that passes through a given triangle, then so does the third: (i) ℓ bisects the area of the triangle; (ii) ℓ bisects the perimeter of the triangle; (iii) ℓ contains the incenter of the triangle.

Verena Haider discovered a beautiful theorem which has the preceding results as immediate consequences.

Haider's Theorem. For any triangle ABC and any line ℓ , ℓ divides the area and the perimeter of ΔABC in the same ratio if and only if it passes through the triangle's incenter.

Proof. We first assume that ℓ divides the area and the perimeter of ΔABC in the same ratio. In our previous notation the ratio of areas is

[EBCD][AED]=[ABC]−[AED][AED]=[ABC][AED]−1,

while the ratio of perimeters is

EB+BC+CDDA+AE=(c−c′)+a+(b−b′)b′+c′.

The right-hand sides are equal when

[ABC][AED]=1+(c−c′)+a+(b−b′)b′+c′=a+b+cb′+c′,

or

[AED]=b′+c′a+b+c[ABC]=b′+c′a+b+c⋅(a+b+c)r2=(b′+c′)r2.

We want to prove that I lies on DE ; to this end we let the bisector of ∠BAC meet DE at F . The perpendicular distances from F to AC and AE have the same value, say d . Therefore,

[AED]=[AEF]+[AFD]=c′d2+b′d2=(b′+c′)d2.

Equations (1) and (2) imply that d=r . Because I is the unique point on the angle bisector AI at a distance of r from AB and AC , it follows that F coincides with I , whence I lies on DE as claimed.



For the converse we are given that I lies on DE . We start the chain of equalities with the ratio of areas,


and end with the ratio of the two pieces of the perimeter; that is, the two ratios are equal as desired.
Further comments.

• Let us call a line that simultaneously bisects the area and perimeter a bisecting line. Campo, Desprez, Fondanaiche, and Harrison all addressed the question of existence. To see that for any triangle there must exist a bisecting line, consider the family of lines DE through I as D moves clockwise about the perimeter of the triangle. Define the function f(D) to equal the quantity obtained by subtracting the area to the left of DE (when looking from D to E ) from the area to the right. It is easy to see that f(D) is continuous; moreover, its value for a particular point D is the negative of the value when D and E have switched places. Somewhere between those two points is a position of D for which f(D)=0 , which is therefore a position where DE bisects the area; because it passes through I we know that it must also bisect the perimeter
  
  
• In fact, as Campo Ruiz and Desprez both showed, every triangle has exactly one, two, or three bisecting lines; no other values are possible. It is not too hard to prove this claim; see, for example, [2] and the references there.
  
  
• In a letter to the editor [6], Anthony Todd wrote that he learned of the bisecting-line problem from a 1994 article by A. Shen [4]. That article describes the discrimination against certain ethnic groups in entrance examinations to the Mekh-mat at Moscow State University during the 1970's and 1980's. One of the difficult problems designed to eliminate many of the targeted applicants was to draw a straight line that bisects the area and perimeter of a triangle. We have seen that showing the existence of such a line is quite easy; constructing that line is much harder. A solution to the harder problem can be found in [7], pages 19-20 and (in Spanish) by Campo Ruiz [1]. Todd's letter concludes by listing several interesting references; one, in particular, is his interactive web page [5] with an applet that displays the bisecting lines of an arbitrary triangle ABC with fixed side AB ; vertex C is free to move about the plane. The applet indicates the regions in which the vertex C must lie in order for one or three bisecting lines to appear, with their common boundary being the locus of C for which two bisecting lines appear.
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